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For given numbers a and b in function aplusb, return the sum of them.
You don‘t need to parse the input and output. Just calculate and return.
If a=1 and b=2 return 3
Can you do it with out + operation?
Are a and b both 32-bit integers?
- Yes.
Analysis:
Use bit manipulation.
Solution 1:
Caluclate every bit.
1 class Solution { 2 /* 3 * param a: The first integer 4 * param b: The second integer 5 * return: The sum of a and b 6 */ 7 public int aplusb(int a, int b) { 8 int res = 0; 9 int carry = 0; 10 for (int i=0;i<32;i++){ 11 int a1 = a & 1; 12 int b1 = b & 1; 13 int val = 0; 14 if (a1==1 && b1==1 && carry==1){ 15 val = 1; 16 carry = 1; 17 } else if ( (a1==1 && b1==1) || (a1==1 && carry==1) || (b1==1 && carry==1) ) { 18 val = 0; 19 carry = 1; 20 } else if (a1==1 || b1==1 || carry==1) { 21 val = 1; 22 carry = 0; 23 } else { 24 val = 0; 25 carry = 0; 26 } 27 val = val << i; 28 res = res | val; 29 a = a >> 1; 30 b = b >> 1; 31 } 32 33 return res; 34 35 } 36 };
Solution 2:
For a + b in any base, we can treat the plus as two part: 1. a + b without carry; 2. the carry generated by a +b. The a+b then equals to part 1 plus part 2. If part1+part2 generates more carry, we can then repeat this procedure, until there is no carry.
1 class Solution { 2 /* 3 * param a: The first integer 4 * param b: The second integer 5 * return: The sum of a and b 6 */ 7 public int aplusb(int a, int b) { 8 while (b!=0){ 9 int carry = a & b; //their carry (actuall, need to move to right by one bit. 10 a = a^b; //their plus result without carry. 11 b = carry << 1; 12 } 13 return a; 14 } 15 };
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原文地址：http://www.cnblogs.com/lishiblog/p/4194937.html
